Meta Hacker Cup 2023 Round 1 题解_hackercup 2023-优快云博客

博客包含多个算法问题。有将数列分组求组中位数最值差的问题；给定正整数 P 找和为 41、积为 P 的数组问题；维护数列操作并求全局最大值位置问题，用线段树解决；还有关于多个字符串的询问问题。

Problem A: Here Comes Santa Claus

给一个数列，要求分成若干组，要求每组至少2个数，使得所有组中位数的最大值与最小值之差尽量大，求这个值。

#include<bits/stdc++.h> 
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,0x3f,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define MEMx(a,b) memset(a,b,sizeof(a));
#define INF (0x3f3f3f3f)
#define F (1000000007)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int> 
#define pi pair<int,int>
#define SI(a) ((a).size())
#define Pr(kcase,ans) printf("Case #%d: %lld\n",kcase,ans);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;
#define PRi2D(a,n,m) For(i,n) {
     
      \
						For(j,m-1) cout<<a[i][j]<<' ';\
						cout<<a[i][m]<<endl; \
						} 
#pragma comment(linker, "/STACK:102400000,102400000")
#define ALL(x) (x).begin(),(x).end()
#define gmax(a,b) a=max(a,b);
#define gmin(a,b) a=min(a,b);
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
ll mul(ll a,ll b){
   
   return (a*b)%F;}
ll add(ll a,ll b){
   
   return (a+b)%F;}
ll sub(ll a,ll b){
   
   return ((a-b)%F+F)%F;}
void upd(ll &a,ll b){
   
   a=(a%F+b%F)%F;}
inline int read()
{
   
   
	int x=0,f=1; char ch=getchar();
	while(!isdigit(ch)) {
   
   if (ch=='-') f=-1; ch=getchar();}
	while(isdigit(ch)) {
   
    x=x*10+ch-'0'; ch=getchar();}
	return x*f;
} 
#define MAXN (100000+10)
ll a[MAXN];

int main()
{
   
   
	freopen("here_comes_santa_claus_input.txt","r",stdin);
	freopen("A.out","w",stdout);
	int T=read();
	For(kcase,T) {
   
   
		printf("Case #%d: ",kcase);
		int n=read();
		For(i,n) a[i]=read();
		sort(a+1,a+1+n);
		if(n==5) {
   
   
			ll p1=(a[2]+a[1]);
			ll p2=(a[5]+a[3]);
			ll p=p2-p1;
			ll p3=(a[3]+a[1]);
			ll p4=(a[5]+a[4]);
			ll _p=p4-p3;
			
			p=max(p,_p);
			if(p%2) cout<<p/2<<".5"<<endl;
			else cout<<p/2<<endl;
		}else